Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 3;
int n, m, q, x, y, zd, l, t, aa, tin[N], tout[N], d[N], up[N][20];
vector<int> a[N], b[N], c[N][20], rez;
vector<int> me(vector<int> u, vector<int> v) {
  vector<int> z;
  int i = 0, j = 0, k = 10;
  int J = v.size();
  int I = u.size();
  if (I == 0) {
    z = v;
  } else if (J == 0) {
    z = u;
  } else {
    while ((i + j < k) && (i + j < I + J)) {
      if (i == I) {
        z.push_back(v[j++]);
      } else if (j == J) {
        z.push_back(u[i++]);
      } else if (u[i] < v[j]) {
        z.push_back(u[i++]);
      } else if (u[i] > v[j]) {
        z.push_back(v[j++]);
      } else {
        ++k;
        ++i;
      }
    }
  }
  return z;
}
void dfs(int v, int p) {
  tin[v] = t++;
  up[v][0] = p;
  d[v] = d[p] + 1;
  for (int i = 1; i <= l; ++i) {
    up[v][i] = up[up[v][i - 1]][i - 1];
    c[v][i] = me(c[v][i - 1], c[up[v][i - 1]][i - 1]);
  }
  for (int u : a[v]) {
    if (u != p) {
      dfs(u, v);
    }
  }
  tout[v] = t++;
}
bool is_parent(int u, int v) { return tin[u] <= tin[v] && tout[u] >= tout[v]; }
void lca(int u, int v) {
  if (is_parent(u, v)) {
    int j = d[v] - d[u];
    for (int i = l; i >= 0; --i) {
      if (j - (1 << i) >= 0) {
        j -= (1 << i);
        rez = me(rez, c[v][i]);
        v = up[v][i];
      }
    }
    rez = me(rez, c[v][0]);
  } else if (!is_parent(v, u)) {
    for (int i = l; i >= 0; --i) {
      if (!is_parent(up[u][i], v)) {
        rez = me(rez, c[u][i]);
        u = up[u][i];
      }
    }
    rez = me(rez, c[u][1]);
  }
}
int main() {
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; ++i) {
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  for (int i = 1; i <= m; ++i) {
    cin >> x;
    c[x][0].push_back(i);
  }
  l = ceil(log2(n));
  a[0].push_back(1);
  dfs(0, 0);
  for (int i = 0; i < q; ++i) {
    cin >> x >> y >> aa;
    if (x != y) {
      rez.clear();
      lca(x, y);
      lca(y, x);
      if (aa > rez.size()) {
        aa = rez.size();
      }
      cout << aa << " ";
      for (int j = 0; j < aa; ++j) {
        cout << rez[j] << " ";
      }
      cout << "\n";
    } else {
      if (aa > c[x][0].size()) {
        aa = c[x][0].size();
      }
      cout << aa << " ";
      for (int j = 0; j < aa; ++j) {
        cout << c[x][0][j] << " ";
      }
      cout << "\n";
    }
  }
  return 0;
}